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College Algebra : Exponential and Logarithmic Functions

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Example Question #2 :Logarithmic Functions

Simplify the following:

$\ln{2x}+\ln{y^2}-\ln{z^7}$

Possible Answers:

$\ln{\frac{y^2z^7}{2x}}$

$\ln{\frac{z^7}{2xy^2}}$

$\ln{\frac{2xz^7}{y^2}}$

$\ln{\frac{2xy^2}{z^7}}$

Correct answer:

$\ln{\frac{2xy^2}{z^7}}$

Explanation:

To solve, you must combine the logs into 1 log, instead of three separate ones. To do this, you must remember that when adding logs, you multiply their insides, and when you subtract them, you add their insides. Therefore,

$\ln{2x}+\ln{y^2}-\ln{z^7}$

$\ln{2xy^2}-\ln{z^7}$

$\ln{\frac{2xy^2}{z^7}}$

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Example Question #3 :Logarithmic Functions

Solve for y in the following expression:

$\log(y) + \log(5x) = 2$

Possible Answers:

y=100-5x

y = 20x

$y = 2 - \log(5x)$

$y = \frac{20}{x}$

Correct answer:

$y = \frac{20}{x}$

Explanation:

To solve for y we first need to get rid of the logs.

$\log(y)+\log(5x) = 2 \rightarrow 10^{\log(y)} \cdot 10^{\log(5x)} = 100$

Then we get $y \cdot 5x = 100$ .

After that, we simply have to divide by 5x on both sides:

$y = \frac{100}{5x} = \frac{20}{x}$

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Example Question #1 :Logarithmic Functions

Solve for.

ln(x^2 -3)=0

Possible Answers:

x={2}

x=-2

$x=\pm{4}$

$x=\pm{\sqrt{2}}$

$x=\pm{2}$

Correct answer:

$x=\pm{2}$

Explanation:

To solve this natural logarithm equation, we must eliminate the ln() operation. To do that, we must remember that ln() is simply log with base. So, we raise both side of the equation to thepower.

$e^{ln(x^2 -3)}=e^0$

This simplifies to

x^2-3=1 . Remember that anything raised to the 0 power is 1.

Continuing to solve for x,

x^2=4

$x=\pm2$

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Example Question #5 :Logarithmic Functions

Solve for.

$log_{7}(7x+7)=1$

Possible Answers:

x=0

x=7

$x=\sqrt{1}$

x=1

x=-7

Correct answer:

x=0

Explanation:

消除 log() operation, simply raise both side of the equation to the 7th 因为基础of the log() operation is 7.

$7^{log_{7}(7x+7)}=7^{1}$

This simplifies to

7x+7=7

7x=0

x=0

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Example Question #6 :Logarithmic Functions

N > 0 ; $a \in (0, 1) \cup (1, \infty)$

True or false:

$\log_{a}N = \frac{\log_{b}N}{\log_{b}a}$

if and only if either b= 10 or b= e .

Possible Answers:

False

True

Correct answer:

False

Explanation:

$\log_{a}N = \frac{\log_{b}N}{\log_{b}a}$

is a direct statement of the Change of Base Property of Logarithms. If N > 0 and $a \in (0, 1) \cup (1, \infty)$ , this property holds true forany $b \in (0, 1) \cup (1, \infty)$ - not just b = e, 10 .

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Example Question #7 :Logarithmic Functions

Evaluate $\log_{-1} (-1)$

Possible Answers:

$\log_{-1} (-1) = 1$ is an undefined quantity.

Correct answer:

$\log_{-1} (-1) = 1$ is an undefined quantity.

Explanation:

$\log_{-1} (-1)$ is undefined for two reasons: first, the base of a logarithm cannot be negative, and second, a negative number cannot have a logarithm.

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Example Question #8 :Logarithmic Functions

Use the properties of logarithms to rewrite as a single logarithmic expression:

$- 2+ t \log 2$

Possible Answers:

$\log \frac{t}{50}$

$\log( 2^{t} -100)$

$\log \frac{2^{t} }{100}$

$\log \frac{t^{2} }{100}$

$\log(2t -100)$

Correct answer:

$\log \frac{2^{t} }{100}$

Explanation:

$b \log M = \log M^{b}$ , so

$- 2+ t \log 2$

$= - 2+ \log 2^{t}$

$N= \log 10^{N}$ , so the above becomes

$= \log 10^{-2}+ \log 2^{t}$

$= \log \frac{1}{100}+ \log 2^{t}$

$\log M + \log N = \log MN$ , so the above becomes

$=\log\left ( \frac{1}{100} \cdot 2^{t} \right )$

$=\log \frac{2^{t} }{100}$

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Example Question #9 :Logarithmic Functions

Use the properties of logarithms to rewrite as a single logarithmic expression:

$\frac{2+ \log_{5}t}{\log_{5}17}$

Possible Answers:

$\log_{17}(32 +t)$

$\log_{17}(25 +t)$

$\log_{17}25 t$

None of the other choices gives the correct response.

$\log_{17}32 t$

Correct answer:

$\log_{17}25 t$

Explanation:

$N= \log_{a} a^{N}$ , so

$\frac{2+ \log_{5}t}{\log_{5}17}$

$= \frac{ \log_{5}5^{2}+ \log_{5}t}{\log_{5}17}$

$= \frac{ \log_{5}25+ \log_{5}t}{\log_{5}17}$

$\log_{a} M + \log_{a} N = \log_{a} MN$ , so the above becomes

$= \frac{ \log_{5}25 t}{\log_{5}17}$

By the Change of Base Property,

$\frac{\log_{a}M}{\log_{a}N} = \log_{N}M$ , so the above becomes

$= \log_{17}25 t$ ,

the correct response.

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Example Question #10 :Logarithmic Functions

Expand the logarithm: $\ln \sqrt[3]{\frac{x}{y}}$

Possible Answers:

$\frac{1}{3} ( \ln x -\ln y)$

$\ln x +3\ln y$

None of these

$\frac{1}{3} ( \ln x +\ln y)$

$3 \ln x -\ln y$

Correct answer:

$\frac{1}{3} ( \ln x -\ln y)$

Explanation:

We expand this logarithm based on the property: $\ln \frac{u}{v}= \ln u-\ln v$

and $\ln u^{n}= n \ln u$ .

$\ln \sqrt[3]{\frac{x}{y}}\rightarrow\frac{1}{3}\ln\frac{x}{y}\rightarrow \boldsymbol{\frac{1}{3}(\ln x-\ln y)}$

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Example Question #11 :Logarithmic Functions

Expand this logarithm: $\ln \frac{x^4\sqrt{y}}{z^6}$

Possible Answers:

$4\ln x+2\ln y-6\ln z$

$4\ln x-\frac{1}{2}\ln y-6\ln z$

$4\ln x+\frac{1}{2}\ln y+6\ln z$

$4\ln x+2\ln y+6\ln z$

$4\ln x+\frac{1}{2}\ln y-6\ln z$

Correct answer:

$4\ln x+\frac{1}{2}\ln y-6\ln z$

Explanation:

We expand this logarithm based on the following properties:

$\ln \frac{u}{v}= \ln u-\ln v$

$\ln u^{n}= n \ln u$

$\ln uv=\ln u+\ln v$

$\ln \frac{x^4\sqrt{y}}{z^6}\rightarrow \ln x^4\ln\sqrt{y}-\ln z^6\rightarrow \boldsymbol{4\ln x+\frac{1}{2}\ln y-6\ln z}$

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College Algebra : Exponential and Logarithmic Functions

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #2 :Logarithmic Functions

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